Problem: You have found the following ages (in years) of all 5 zebras at your local zoo: $ 10,\enspace 30,\enspace 21,\enspace 13,\enspace 13$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{10 + 30 + 21 + 13 + 13}{{5}} = {17.4\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $10$ years $-7.4$ years $54.76$ years $^2$ $30$ years $12.6$ years $158.76$ years $^2$ $21$ years $3.6$ years $12.96$ years $^2$ $13$ years $-4.4$ years $19.36$ years $^2$ $13$ years $-4.4$ years $19.36$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{54.76} + {158.76} + {12.96} + {19.36} + {19.36}} {{5}} $ $ {\sigma^2} = \dfrac{{265.2}}{{5}} = {53.04\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{53.04\text{ years}^2}} = {7.3\text{ years}} $ The average zebra at the zoo is 17.4 years old. There is a standard deviation of 7.3 years.